The distance between the parallel lines 9x^2 | vedclass

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(D) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$. We can rewrite the quadratic part as $(3x - y)^2 + 6(3x - y) + 8 = 0$. Let $t = 3x -

Vedclass · Accepted Answer

$\frac{2}{\sqrt{10}}$

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(D) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$. We can rewrite the quadratic part as $(3x - y)^2 + 6(3x - y) + 8 = 0$. Let $t = 3x - y$. Then the equation becomes $t^2 + 6t + 8 = 0$. Factoring the quadratic,we get $(t + 4)(t + 2) = 0$. Thus,the two parallel lines are $3x - y + 4 = 0$ and $3x - y + 2 = 0$. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here,$a = 3$,$b = -1$,$c_1 = 4$,and $c_2 = 2$. $d = \frac{|4 - 2|}{\sqrt{3^2 + (-1)^2}} = \frac{2}{\sqrt{9 + 1}} = \frac{2}{\sqrt{10}}$.

The distance between the parallel lines $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$ is

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